First derivative of the legendre polynomials

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SUMMARY

The first derivative of Legendre polynomials satisfies a self-adjoint differential equation characterized by the eigenvalue λ = n(n + 1) - 2. The equation is expressed as (1 - x^2) P_n'' - 2x P_n' = λ P_n. For n = 2, the equation simplifies to -15x^2 + 2x + 3 = 0, demonstrating the relationship between the derivatives of Legendre polynomials and their eigenvalues. The discussion emphasizes the importance of differentiating the polynomial and substituting variables to simplify the equation.

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  • Understanding of Legendre polynomials
  • Knowledge of differential equations
  • Familiarity with eigenvalues and eigenfunctions
  • Basic calculus, including differentiation techniques
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  • Study the properties of Legendre polynomials in detail
  • Learn about self-adjoint differential equations
  • Explore the method of solving eigenvalue problems in differential equations
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Mathematicians, physicists, and students studying advanced calculus or differential equations, particularly those interested in the properties and applications of Legendre polynomials.

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show that the first derivative of the legendre polynomials satisfy a self-adjoint differential equation with eigenvalue λ=n(n+1)-2


The attempt at a solution:

(1-x^2 ) P_n^''-2xP_n^'=λP_n
λ = n(n + 1) - 2 and (1-x^2 ) P_n^''-2xP_n^'=nP_(n-1)^'-nP_n-nxP_n^'
∴nP_(n-1)^'-nP_n-nxP_n^'=( n(n + 1) - 2 )P_n
For n=2
2P_1^'-2P_2-2xP_2^'=4P_n
Using legendre polynomial properties we have
-9x^2+2x+1=2(3x^2-1)
∴-15x^2+2x+3=0
I don’t know what to do from here.
 
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Did you consider just replacing the first derivative with just "y" so that the equation becomes
(1- x^2)y'- 2xy= \lambda P_n
Differentiating
(1- x^2)y''- 4xy'- 2y= \lambda P_n'= \lambda y

(1- x^2)y''- 4xy'= (2+ \lambda)y
 
Thanks Hallsofvy
 

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